Polynomial Functions, Zeros, Factors, and Intercepts

How zeros, factors, and x-intercepts connect — with 4 fully solved problems

Overview

This tutorial reviews the connections between zeros, factors, x-intercepts, and multiplicities of polynomial functions, then works through four practice problems with full solutions — from reconstructing a cubic from its graph to finding all zeros of a quartic given one complex root.

Review: Properties of Zeros and Factors

A) Real Zeros

Let \( p(x) \) be a polynomial function with real coefficients. If \( p(s) = 0 \), the following statements are mathematically equivalent:

B) Multiplicity — touch vs. cross

A zero's multiplicity is the exponent of its linear factor in the fully factored form of the polynomial.

C) Complex Zeros

The imaginary unit is defined by \( i^2 = -1 \). If \( p(x) \) has real coefficients and \( a + bi \) (with \( b \ne 0 \)) is a zero of \( p(x) \), then its complex conjugate \( a - bi \) is also a zero. This follows from conjugating \( p(a + bi) = 0 \) and using the fact that real coefficients are fixed by conjugation.

Example — Connecting the properties

Let \( p(x) = x^3 - 2x^2 + 2x - 4 \).

  1. \( p(2) = 2^3 - 2(2^2) + 2(2) - 4 = 8 - 8 + 4 - 4 = 0 \), so \( 2 \) is a zero of \( p(x) \).
  2. \( x = 2 \) is a solution of the equation \( p(x) = 0 \).
  3. \( p(x) \) factors as \( p(x) = (x - 2)(x^2 + 2) \).
  4. The graph of \( p(x) \) has an x-intercept at \( x = 2 \).
Graph of y = x³ − 2x² + 2x − 4 with a single x-intercept at x = 2
Graph of \( p(x) = x^3 - 2x^2 + 2x - 4 \).

Practice Problems with Solutions

Problem 1 — Reconstruct a cubic from its graph

The graph below represents a polynomial function \( p(x) \) with real coefficients. The degree of \( p(x) \) is 3, and all zeros are integers. Find the equation for \( p(x) \).

Graph of a cubic with an x-intercept at x = 2, a turning-point x-intercept at x = −1, and a y-intercept at (0, −2)
Graph for Problem 1 — degree-3 polynomial with two x-intercepts and y-intercept \( (0, -2) \).
Solution

Step 1 — Identify the x-intercepts and their multiplicities. The graph shows:

  • \( x = -1 \) — where the graph touches the x-axis and turns around (even multiplicity, so 2 here since the degree is 3 and there's another intercept);
  • \( x = 2 \) — where the graph crosses the x-axis (odd multiplicity, so 1).

Since the multiplicities must add up to the degree (3), the zero at \( -1 \) has multiplicity exactly 2. The general form is:

\[ p(x) = a(x + 1)^2(x - 2), \qquad a \ne 0 \]

Step 2 — Use the y-intercept to find \( a \). The y-intercept is \( (0, -2) \), so \( p(0) = -2 \):

\[ a(0 + 1)^2(0 - 2) = -2 \] \[ -2a = -2 \implies a = 1 \]

Final answer:

\[ p(x) = (x + 1)^2 (x - 2) \]

Check: \( p(0) = (1)^2(-2) = -2 \) ✓; \( p(-1) = 0 \) (double root) ✓; \( p(2) = 0 \) ✓

Problem 2 — Degree-5 polynomial from its zeros

A polynomial function \( p(x) \) of degree 5 with real coefficients has the zeros \( -1 \), \( 2 \) (with multiplicity 2), \( 0 \), and \( 1 \). If \( p(3) = -12 \), find the equation for \( p(x) \).

Solution

Step 1 — Write the factored form. Degrees add up: \( 1 + 2 + 1 + 1 = 5 \) ✓, so:

\[ p(x) = a\,x\,(x + 1)\,(x - 2)^2\,(x - 1), \qquad a \ne 0 \]

Step 2 — Use \( p(3) = -12 \) to solve for \( a \).

\[ a(3)(3 + 1)(3 - 2)^2(3 - 1) = -12 \] \[ a \cdot 3 \cdot 4 \cdot 1 \cdot 2 = -12 \] \[ 24a = -12 \implies a = -\tfrac{1}{2} \]

Final answer:

\[ p(x) = -\tfrac{1}{2}\,x\,(x + 1)\,(x - 2)^2\,(x - 1) \]

Check: \( p(3) = -\tfrac{1}{2} \cdot 3 \cdot 4 \cdot 1 \cdot 2 = -12 \) ✓

Problem 3 — Find all zeros from one complex zero

The complex number \( 2 + i \) is a zero of \( p(x) = x^4 - 2x^3 - 6x^2 + 22x - 15 \). Find all other zeros.

Solution

Step 1 — Use the conjugate-root theorem. Because the coefficients are real, the conjugate \( 2 - i \) is also a zero. Hence \( (x - (2+i)) \) and \( (x - (2-i)) \) are factors.

Step 2 — Multiply to get a real quadratic factor.

\[ (x - (2+i))(x - (2-i)) = (x - 2)^2 - i^2 = x^2 - 4x + 4 + 1 = x^2 - 4x + 5 \]

Step 3 — Divide the original polynomial by this quadratic.

\[ q(x) = \frac{x^4 - 2x^3 - 6x^2 + 22x - 15}{x^2 - 4x + 5} = x^2 + 2x - 3 \]

Verify: \( (x^2 - 4x + 5)(x^2 + 2x - 3) = x^4 - 2x^3 - 6x^2 + 22x - 15 \) ✓

Step 4 — Factor \( q(x) \).

\[ x^2 + 2x - 3 = (x - 1)(x + 3) \]

Final answer — all zeros: \( 2 + i,\; 2 - i,\; 1,\; -3 \)

Problem 4 — Another complex-zero factorization

The complex number \( -3 - i \) is a zero of \( p(x) = x^4 + 6x^3 + 11x^2 + 6x + 10 \). Find all other zeros.

Solution

Step 1 — Conjugate root. The conjugate \( -3 + i \) is also a zero, so \( (x + 3 + i) \) and \( (x + 3 - i) \) are factors.

Step 2 — Multiply to get a real quadratic factor.

\[ (x + 3 + i)(x + 3 - i) = (x + 3)^2 - i^2 = x^2 + 6x + 9 + 1 = x^2 + 6x + 10 \]

Step 3 — Divide.

\[ q(x) = \frac{x^4 + 6x^3 + 11x^2 + 6x + 10}{x^2 + 6x + 10} = x^2 + 1 \]

Verify: \( (x^2 + 6x + 10)(x^2 + 1) = x^4 + 6x^3 + 11x^2 + 6x + 10 \) ✓

Step 4 — Solve \( q(x) = 0 \).

\[ x^2 + 1 = 0 \implies x = \pm i \]

Final answer — all zeros: \( -3 - i,\; -3 + i,\; i,\; -i \)

Try These Yourself

Practice 1 — Reconstruct a quadratic

A degree-2 polynomial has a zero at \( x = 3 \) (with multiplicity 2) and passes through \( (0, 18) \). Find its equation.

Answer

\( p(x) = a(x - 3)^2 \). From \( p(0) = 18 \): \( a(0-3)^2 = 9a = 18 \Rightarrow a = 2 \).

So \( p(x) = 2(x - 3)^2 \).

Practice 2 — Find all zeros given one complex zero

The complex number \( 1 + i \) is a zero of \( p(x) = x^3 - 3x^2 + 4x - 2 \). Find all zeros.

Answer

Conjugate \( 1 - i \) is also a zero. Product: \( (x - (1+i))(x - (1-i)) = (x-1)^2 + 1 = x^2 - 2x + 2 \).

Divide: \( p(x) \div (x^2 - 2x + 2) = x - 1 \).

All zeros: \( 1 + i,\; 1 - i,\; 1 \).

Practice 3 — Identify multiplicities from a graph

A degree-4 polynomial \( p(x) \) touches the x-axis at \( x = 0 \) and crosses it at \( x = 3 \). All zeros are integers. What are the possible multiplicities?

Answer

Touching at \( x = 0 \) → even multiplicity (2 or 4). Crossing at \( x = 3 \) → odd multiplicity (1 or 3). The sum must be 4.

Possible combinations: \( (2, 2) \) — i.e., multiplicity 2 at each. (Multiplicity 4 at \( x=0 \) would leave none at \( x=3 \); multiplicity 3 at \( x=3 \) would leave 1 for \( x=0 \), which is odd — impossible for a "touching" behavior.)

So multiplicities are 2 at \( x=0 \) and 2 at \( x=3 \).

Frequently Asked Questions

What is the relationship between zeros, factors, and x-intercepts of a polynomial?

For a polynomial \(p(x)\) with real coefficients, the following are equivalent: \(s\) is a zero (\(p(s) = 0\)); \(s\) is a solution of \(p(x) = 0\); \((x - s)\) is a factor of \(p(x)\); and \((s, 0)\) is an x-intercept of the graph.

Why do complex zeros come in conjugate pairs?

If \(p(x)\) has real coefficients and \(z = a + bi\) is a zero, conjugating \(p(z) = 0\) gives \(p(a - bi) = 0\) as well — because conjugation commutes with real-coefficient arithmetic. So \(a - bi\) is also a zero.

How do you reconstruct a polynomial from its zeros?

Write \(p(x) = a \cdot (x - s_1)^{m_1} \cdot (x - s_2)^{m_2} \cdots\) where \(s_i\) are the zeros and \(m_i\) are their multiplicities. The unknown leading coefficient \(a\) is determined by one additional data point — usually a y-intercept or another point on the graph.

What does it mean for a graph to "touch" the x-axis without crossing?

A zero with even multiplicity (2, 4, …) causes the graph to touch the x-axis and turn around — the sign of the polynomial doesn't change. A zero with odd multiplicity (1, 3, …) causes the graph to cross.

More Tutorials and References on Polynomial Functions

Last updated: October 17, 2026